TMJC Chp8 Applications of Differentiation Assignment Solutions

Chapter 8 Applications of Differentiation TMJC 2024 H2 Mathematics (9758) Chapter 8 Applications of Differentiation Assignment Suggested Solutions 1 2019/SAJC Promo/Q8 The curve C has parametric equations   23 for 03 , ln , 3.t y t tx      (i) Sketch the graph of C, giving the coordinates of its endpoint(s) and the point(s) where C meets the axes. State also the equation of the vertical asymptote. [3] (ii) Find the equation of the tangent to the curve C at the point   3 2 , ln3 p p        , simplifying your answer. [5] (iii) Hence find the exact coordinates of the points Q and R where the tangent to the curve C when et  meets the x-axis and y-axis respectively. [3] (iv) Find the area of triangle OQR in exact form. [2] Q1 Solution (i) As 0, , 0.yt x   Hence vertical asymptote is 0.x  Intercept: 11 30, t xy    x O y (9, [ln(3)]ଶ) ൬1 3 , 0൰ ݔ= 0 Remember to press p and set Tmin = 0 Tmax = 3 For sketch, need to label the endpoint as a closed circle, since endpoint is at t = 3 which is included. Question asks to state equation of vertical asymptote. Keep this in mind and make sure you have a vertical asymptote in your graph. As 0t  , 3 03x t  ln t ,   2 lny t  Asymptote: 0x

Chapter 8 Applications of Differentiation TMJC 2024 (ii) 2 2d 3 d 2ln ,d d 3 x t y t tt t t   2 3 d d 1 dd d d 2 2l ln 1 n y y xx t t t t t t t      At ,t p 3 2lnd .dx py p Equation of the tangent of C at point p :       3 2 3 2 3 2 3 2lnln 3 2 2ln ln ln 3 2 2ln ln ln 3 p py p x p y p p x p p y p x p pp                     ALERT:  2ln 2lnp p  (iii) Equation of the tangent of C at point t = p :   2 3 2 2ln ln ln 3 y p x p p p       At ep ,   2 3 3 2 2ln e ln e ln ee 3 2 1 e 3 y x y x         When the tangent cuts the x – axis, 3e0 . 6y x   When the tangent cuts the axis at y – axis, 1.0 3yx   The coordinates of Q are 3e .6 , 0    The coordinates of R are 10, .3      Note: t is the parameter, p is a constant that gives a specific point So, you have to differentiate with respect to t, then substitute t = p to find the value of the gradient at that point Keyword: ‘Hence’ Means you need to use the previous result (equation of tangent at point t = p) to solve Reminder: Question wants exact coordinates so remember to answer the question Recall: Equation of line: ݕ−ݕ଴ =݉(ݔ−ݔ଴) where m is the gradient and (ݔ଴,ݕ଴) is a point on the line You need to find the value of the gradient at point P by substituting t p Note: e is a constant Use equation of tangent found in (ii), substituting ep . Simplify using ln e 1

Chapter 8 Applications of Differentiation TMJC 2024 (iv) 3 3 2 1 e 1Area of triangle 2 6 3 e units36 OQR            2 2019/JPJC Promo/Q4 Steel sheets of negligible thickness are use